x/(x-√(x^2-1)) dx is equal to : Option 1: x^3 + ((x^2-1)^(3/2))/3 + C,where C is an arbitrary constant. Option 2: x^4/3 + ((x^2-1)^(3/2))/3 + C,where C is an arbitrary constant. Option 3: x^3/3 - ((x^2-1)^(3/2))/3 + C,where C is an arbitrary constant. Option 4: x^3/3 + ((x^2-1)^(3/2))/3 + C,where C is an arbitrary constant.

Answer: x^3/3 + ((x^2-1)^(3/2))/3 + C,where C is an arbitrary constant. Solution: Rationalise by multiplying numerator and denominator by x+√(x^2-1); the denominator becomes x^2-(x^2-1) = 1. The integral reduces to x(x+√(x^2-1))dx = x^2 dx + x√(x^2-1) dx. This gives x^3/3+((x^2-1)^(3/2))/3+C.

Calculus

Integrals

Medium

applied

$\displaystyle\int \dfrac{x}{x-\sqrt{x^2-1}}\,dx$ is equal to :

$x^3 + \dfrac{(x^2-1)^{3/2}}{3} + C$ ,where $C$ is an arbitrary constant.

$\dfrac{x^4}{3} + \dfrac{(x^2-1)^{3/2}}{3} + C$ ,where $C$ is an arbitrary constant.

$\dfrac{x^3}{3} - \dfrac{(x^2-1)^{3/2}}{3} + C$ ,where $C$ is an arbitrary constant.

$\dfrac{x^3}{3} + \dfrac{(x^2-1)^{3/2}}{3} + C$ ,where $C$ is an arbitrary constant.

✅ Correct Option: 4

Rationalise by multiplying numerator and denominator by $x+\sqrt{x^2-1}$ ; the denominator becomes $x^2-(x^2-1) = 1$ .

The integral reduces to $\int x\left(x+\sqrt{x^2-1}\right)dx = \int x^2 dx + \int x\sqrt{x^2-1}\,dx$ .

This gives $\dfrac{x^3}{3}+\dfrac{(x^2-1)^{3/2}}{3}+C$ .

2026: 12 May Shift 2

Match List-I with List-II

List-I	List-II
Indefinite integral	Solution (where 'c' is an arbitrary constant)
(A) $\displaystyle\int \sqrt{x^2-a^2} \, dx$	(I) $\dfrac{1}{a}\tan^{-1}\dfrac{x}{a} + c$
(B) $\displaystyle\int \dfrac{1}{x^2-a^2} \, dx$	(II) $\dfrac{x}{2}\sqrt{x^2+a^2} + \dfrac{a^2}{2}\log\left[x+\sqrt{x^2+a^2}\right] + c$
(C) $\displaystyle\int \dfrac{1}{x^2+a^2} \, dx$	(III) $\dfrac{x}{2}\sqrt{x^2-a^2} - \dfrac{a^2}{2}\log\left[x+\sqrt{x^2-a^2}\right] + c$
(D) $\displaystyle\int \sqrt{x^2+a^2} \, dx$	(IV) $\dfrac{1}{2a}\log\left\vert \dfrac{x-a}{x+a} \right\vert + C$

Choose the correct answer from the options given below:

2026: 12 May Shift 2

Match the rational functions with the correct form of the partial fractions.

[ $p,q,r,a,b,c,A,B,C$ are arbitrary constants]

Match List-I with List-II

List-I (Rational Function)	List-II (Partial Fraction)
(A) $\dfrac{px^2+qx+r}{(x-a)(x^2+bx+c)}$	(I) $\dfrac{A}{x-a}+\dfrac{B}{(x-a)^2}+\dfrac{C}{x-b}$
(B) $\dfrac{px^2+qx+r}{(x-a)^2(x-b)}, a\neq b$	(II) $\dfrac{A}{x-a}+\dfrac{B}{(x-a)^2}$
(C) $\dfrac{px^2+qx+r}{(x-a)(x-b)(x-c)}, a\neq b\neq c$	(III) $\dfrac{A}{x-a}+\dfrac{Bx+C}{x^2+bx+c}$
(D) $\dfrac{px+q}{(x-a)^2}$	(IV) $\dfrac{A}{x-a}+\dfrac{B}{x-b}+\dfrac{C}{x-c}$

Choose the correct answer from the options given below:

2025: 22 May Shift 2

$\int \frac{\sin 2x \, dx}{\sqrt{9 - \cos^4 x}}$ equals

2026: 30 May Shift 1

$\int\frac{x-1}{x(x-\log x)}dx$ equals:

(consider: $\log x=\log_e x$ )

2025: 21 May Shift 2

The value of $\int_1^3 \frac{x^2}{x^3+1}dx$

2026: 19 May Shift 1

Match List-I with List-II

[c is an arbitrary constant]

List-I	List-II
(A) $\int \frac{\sin^2 x - \cos^2 x}{\sin^2 x \cos^2 x}\,dx$	(I) $\tan x - x + c$
(B) $\int \frac{\sec^2 x}{\text{cosec}^2 x}\,dx$	(II) $2\tan x - 3\sec x + c$
(C) $\int \sec x(\sec x + \tan x)\,dx$	(III) $\tan x + \cot x + c$
(D) $\int \frac{2-3\sin x}{\cos^2 x}\,dx$	(IV) $\tan x + \sec x + c$

Choose the correct answer from the options given below:

∫xx−x2−1 dx\displaystyle\int \dfrac{x}{x-\sqrt{x^2-1}}\,dx∫x−x2−1​x​dx is equal to :

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

$\displaystyle\int \dfrac{x}{x-\sqrt{x^2-1}}\,dx$ is equal to :